To solve for the electric field strength $E$ in the dielectric of a capacitor, given the polarization $P = 2 \, \mu C/m^2$, relative permittivity $\varepsilon_r = 5.5$, and permittivity of vacuum $\varepsilon_0 = 8.854 \times 10^{-12} \, F/m$, we can use the following relation:

$P = \varepsilon_0 (\varepsilon_r - 1) E$

Where:

• $P$ is the polarization,
• $\varepsilon_0$ is the vacuum permittivity,
• $\varepsilon_r$ is the relative permittivity of the dielectric, and
• $E$ is the electric field strength.

We want to solve for $E$:

$E = \frac{P}{\varepsilon_0 (\varepsilon_r - 1)}$

Now, let's plug in the values:

• $P = 2 \times 10^{-6} \, C/m^2$
• $\varepsilon_0 = 8.854 \times 10^{-12} \, F/m$
• $\varepsilon_r = 5.5$

$E = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12} \times (5.5 - 1)}$

$E = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12} \times 4.5}$

Now let's calculate $E$.The electric field strength $E$ in the dielectric is approximately $50,197 \, \text{V/m}$.

Would you like more details or have any questions about this solution?

Here are five related questions for further exploration:

1. How does the electric field strength change with varying polarization $P$?
2. What effect does increasing the relative permittivity $\varepsilon_r$ have on the electric field strength?
3. How does the electric field strength compare in a vacuum versus in a dielectric?
4. Can you calculate the electric field strength if the relative permittivity is doubled?
5. What is the relationship between capacitance and the electric field in a dielectric material?

Tip: The electric field inside a dielectric material is always weaker than in a vacuum, due to the dielectric's ability to polarize and reduce the field strength.